Mechanical Measurements

Mechanical Measurements Prof. S.P.Venkateshan

Indian Institute of Technology Madras

Sub Module 2.11

Pressure transducers Here we discuss several pressure transducers that are grouped together, for

convenience. These are:

1. Pressure tube with bonded strain gage

2. Diaphragm/Bellows type transducer

3. Capacitance pressure transducer

The common feature of all these transducers is that the pressure to be

measured introduces a strain or movement in a mechanical element that is

measured by different techniques. The strain is measured by a) strain gage

or b) linear voltage displacement transducer (LVDT) or c) the change in

capacitance. Detailed discussion of each one of these follows.

1) Pressure tube with bonded strain gage

The principle of a pressure tube with bonded strain gage may be understood

by referring to Figure 72.

Figure 72 Pressure tube with bonded strain gage

The pressure to be measured is communicated to the inside of a tube of

suitable wall thickness one end of which is closed with a thick plate. Because

p Dummy Gauge

Active Gauge

1. Active gauge responds to hoop strain due to pressure

2. Dummy gauge helps in temperature compensation



the end plate is very thick it undergoes hardly any strain. The tube, however,

experiences a hoop strain due to internal pressure (indicated by the blue

arrows). A strain gage mounted on the tube wall experiences the hoop strain.

The way the strain gage works is dealt with below.

Strain gage theory: Strain gage consists of a resistance element whose resistance is a function of

its size.

Figure 73 Deformation of a wire element

Electrical resistance of a wire of Length L, area of cross section A and made

of a material of specific resistance ρ is given by

LRAρ

= (72)

Logarithmic differentiation yields

d R d d L d AR L A

ρρ

= + − (73)

Equation 73 represents the fractional change in the wire resistance due to

fractional changes in the specific resistance, the length and the area of cross

section due to an imposed deformation of the wire. The specific resistance is

a strong function of temperature but a weak function of pressure. If we

assume that the element is at a constant temperature the first term on the

A-∆A

∆L L

Original shape Changed shape



right hand side of Equation 73 may be dropped to get the relation

L o n g i t u d i n a l F r a c t i o n a lS t r a i n C h a n g e i n a r e a

d R d L d AR L A

= − (75)

If the wire is of circular cross section of radius r, we have 2A rπ= and hence

2

Fractional LateralChange in area Strain

dA drA r

= (76)

Recall from mechanics of solids that Poisson ratio ν for the material is defined

as the ratio of lateral strain to longitudinal strain. Hence we have

2 2dA dr dLA r L

ν= = − (77)

Combining these, Equation 75 becomes

( ) ( )1 2 1 2dR dLR L

ν ν= + = + ∈ (78)

In the above the longitudinal strain is represented by∈. An incompressible

material has Poisson ratio of 0.5. In that case the fractional change in

resistance is equal to twice the longitudinal strain. The ratio of fractional

resistance change of the wire to the longitudinal strain is called the gage

factor (GF) or sensitivity of a strain gage and is thus given by

( )1 2dR

RGF ν= = +∈

(79)

The gage factor is thus close to 2. Poisson ratio of some useful materials is

given in Table 8.



Table 14 Poisson ratio of some useful materials

Table 15 presents typical gage factor values with different materials. Strain gage construction details: Figure 74 shows the way a strain gage is made. The strain element is a

serpentine metal layer obtained by etching, mounted on a backing material

that may is bonded on to the surface whose strain needs to be measured.

The active direction is as indicated in the figure. When a force is applied

normal to this direction the serpentine metal layer opens up like a spring and

hence does not undergo any strain in the material (actually the response will

be something like 1% of that along the active direction). When the length of

the metal foil changes the resistance changes and the change in resistance is

the measured quantity. Connecting wires are used to connect the strain gage

to the external circuit.



Table 15 Typical gage materials and gage factors

Material Sensitivity (GF) Platinum (Pt 100%) 6.1 Platinum-Iridium (Pt 95%, Ir 5%) 5.1 Platinum-Tungsten (Pt 92%, W 8%) 4.0 Isoelastic (Fe 55.5%, Ni 36% Cr 8%, Mn 0.5%) * 3.6

Constantan / Advance / Copel (Ni 45%, Cu 55%) *

2.1

Nichrome V (Ni 80%, Cr 20%) * 2.1 Karma (Ni 74%, Cr 20%, Al 3%, Fe 3%) * 2.0 Armour D (Fe 70%, Cr 20%, Al 10%) * 2.0 Monel (Ni 67%, Cu 33%) * 1.9 Manganin (Cu 84%, Mn 12%, Ni 4%) * 0.47 Nickel (Ni 100%) -12.1 * Isoelastic, Constantan, Advance, Copel, Nichrome V, Karma, Armour D, Monel, and Manganin are all trade marks

Figure 74 Strain gage construction schematic (not to scale)

Connecting wires

Backing material

Etched metal foil

Strain Direction (active)



Figure 75 Strain gage elements as they actually appear (Visit: www.omega.com)

Strain gage elements as supplied by Omega are shown in Figure 75. By

making the element in the form of a serpentine foil the actual length of the

element is several times the length of the longer side of the element (referred

to as gage length in Figure 59). The gage length may vary from 0.2 to 100

mm. For a given strain the change in length will thus be the sum of the

changes in length of each leg of the element and hence the resistance

change will be proportionately larger. Gages with nominal resistances of

120,350,600 and 700 � are available. However 120 � gages are very

common.

Gage length



Example 25

A typical foil strain gage has the following specifications:

1. Material is Constantan 6 �m thick

2. Resistance 120 � ± 0.33%

3. Gage factor 2 ± 10%

Such a strain gage is bonded on to a tube made of stainless steel of

internal diameter 6 mm and wall thickness 0.3 mm. The tube is

subjected to an internal pressure of 0.1 MPa gage. The outside of the

tube is exposed to the standard atmosphere. What is the change in

resistance of the strain gage if the Young’s modulus of stainless steel is

207 GPa?

Data: Applied internal pressure is 50.1 10p MPa Pa= =

Tube internal diameter is 6 0.006ID mm m= = Tube wall thickness is 40.3 3 10t mm m−= = ×

Tube is assumed to be thin and the radius r is taken as the

mean radius given by

( ) ( )0.006 0.00030.00315

2 2ID t

r m+ +

= = =

The Young’s modulus of stainless steel is given to be

8207 2.07 10E GPa Pa= = ×

The hoop stress due to internal pressure is calculated as

5610 0.00315 1.05 10

0.0003hpr Pa Pat

σ ×= = = ×

The corresponding strain is calculated as

6

8

1.05 10 0.00512.07 10

hh E

σ ×∈ = = =

×

(For Constantan the maximum allowable elongation is 1% and the above

value is acceptable)



With the gage factor of 2GF = the change in resistance as a fraction of

the resistance of the strain element is

2 0.0051 0.0101hR GF

RΔ

= ∈ = × =

The change in resistance of the strain gage is thus given by

120 2 0.0051 1.02hR RGFΔ = ∈ = × × = Ω



Bridge circuits for use with strain gages: While discussing resistance thermometers we have already seen how a

bridge circuit may be used for making measurements (see Sub Module 2.3).

Since a strain gage is similar in its operation to an RTD it is possible to use a

similar bridge circuit for measuring the strain and hence the pressure. Lead

wire compensation is also possible by using a three wire arrangement. For

the sake of completeness some of these will be discussed again here.

The basic DC bridge circuit is shown in Figure 76. The output voltage Vo is

related to the input voltage Vs by the relation (derived using Ohm’s law)

( )( )( )

1 3 2 4

1 4 2 3o s

R R R RV V

R R R R−

=+ +

(80)

Figure 76 Basic bridge circuit and the nomenclature If all the resistances are chosen equal (equal to 120 �, for example) when the

strain gage is unstrained, the output voltage will be zero. Let us assume that

R3 is the strain gage element and all others are standard resistors. When the

strain gage is strained the resistance will change to 3 3R R+ Δ where 3 3R RΔ .

The numerator of Equation 80 then becomes 1 3R RΔ . The denominator is

approximated by ( )( ) ( )( ) 21 4 2 3 3 1 4 2 3 14R R R R R R R R R R+ + + Δ ≈ + + = since all

resistances are equal when the strain gage is unstrained. We thus see that

Equation 80 may be approximated by

Vs = supply voltage Vo + -

R1

R4

R2

R3



1 3 3 3

21 1 34 4 4 4o s s s s

R R R R GFV V V V VR R RΔ Δ Δ ∈

= = = = (81)

Thus the output voltage is proportional to the strain experienced by the strain

gage. For a gage with sensitivity of 2 the ratio of output voltage to input

voltage is equal to half the strain. In practice Equation 81 is rearranged to

read

4 1o

s

VV GF

∈= (82)

Since the pressure is proportional to the strain, Equation 82 also gives the

pressure within a constant factor. Note that Vo needs to be measured by the

use of a suitable voltmeter, likely a milli-voltmeter. In practice one may use an

amplifier to amplify the signal by a known factor and measure the amplified

voltage by a suitable voltmeter. Another method is to make the resistance R2

a variable resistor and null the bridge by using a controller as shown in Figure

77.

Figure 77 Null balanced bridge

The controller uses the amplified off balance voltage to drive the bridge to

balance by varying the resistor R2. The voltage Vo shown is then a measure

of the strain.

R1

R4

R2

R3

Controller

Vo

Amplifier



Example 26

The supply voltage in the arrangement depicted in Figure 60 is 5 V. All the

resistors are equal when the strain gage is not under load. When the strain

gage is loaded along its axis the output voltage registered is 1.13 mV. What

is the axial strain of the gage if the sensitivity is 2?

We use Equation 82 for calculating the axial strain. We have:

35 , 1.13 1.13 10 2s oV V V mV V and GF−= = = × = The axial strain is then given by

344 1 4 1.13 10 4.52 10 452

5 2o

as

V micro strainV GF

−−× ×

∈ = = = × = −×

Example 27

Consider the data given in Example 21 again. This gage is connected

in a bridge circuit with an input voltage of 9 V. What is the output

voltage in this case?

120 2 0.0101 1.02hR RGFΔ = ∈ = × × = Ω From the results in Example 21 the change is resistance has been

obtained as 3 1.02RΔ = Ω . We then have 3

3

1.02 0.0085120

RRΔ

= = . With

9sV V= we then have

3

3

1.029 0.0191 19.14 4 120o s

RV V V mVR

Δ= = × = =

×



Three wire gage for lead wire compensation:

Figure 78 Three wire arrangement for lead wire compensation

The connections are made as shown in Figure 78 while using a strain gage

with three leads. Let each lead have a resistance of RL as shown. Also let

'2 2 LR R R= + and '

3 3 LR R R= + . Equation 80 is recast as

( )( )( )

( )( )( )( )( )

' '1 3 2 4

' '1 4 2 3

1 3 2 4

1 4 2 3

1 3 2 4 1 4

1 4 2 3

( ) ( )2

( )2

o s

L Ls

L

Ls

L

R R R RV V

R R R R

R R R R R RVR R R R R

R R R R R R RV

R R R R R

−=

+ +

+ − +=

+ + +

− + −=

+ + +

(83)

If all the resistances are identical under no strain condition the second term in

the numerator is zero and hence the balance condition is not affected by the

lead resistance. Let 3 3R R+ Δ be the resistance in the strained condition for

the gage. Again we assume that 3 3R RΔ and, in addition, we assume

that 3 LR RΔ . With these Equation 83 may be approximated by the relation

( )( )( )

1 3 2 4 1 3 3 3

1 4 2 3 3 11

1

12 4

4 1

Lo s s s

L L

R R R R R R R R RV V V VR R R R R R RRR

R

− + Δ ⎛ ⎞Δ Δ= ≈ ≈ −⎜ ⎟+ + + ⎛ ⎞ ⎝ ⎠+⎜ ⎟

⎝ ⎠

(84)

In writing the above we have additionally assumed that the lead resistance is

small compared to the resistance of the gage. Thus Equation 81 is recast as

1

14

Lo s

RGFV VR

⎛ ⎞∈≈ −⎜ ⎟

⎝ ⎠ (85)

Vo

R3

Vs = supply voltage

R4

R1 R2 RL

RL



Thus the three wire arrangement provides lead wire compensation but with a

reduced effective gage factor of1

1 LRGFR

⎛ ⎞−⎜ ⎟

⎝ ⎠.

Example 28

A three wire strain gage with a lead resistance of 4LR = Ω and a gage

resistance under no strain of 120R = Ω is used to measure a strain of 0.001.

What will be the output voltage for an input voltage of 5 V if a) the lead

resistance effect is not included b) if the lead wire resistance is included?

The gage factor GF has been specified to be 1.95.

Case a

The effective gage factor in this case is the same as GF = 1.9. The output

voltage with a strain of ∈ = 0.001 is given by

31.9 105 0.00238 23.84 4o s

GFV V V mV−∈ ×

≈ = × = =

Case b

The lead resistance is given to be 4 �. The resistance of the gage is R = 120

�. Hence the effective gage factor is

41 1.9 1 1.837120

Leff

RGF GFR

⎛ ⎞ ⎛ ⎞= − = × − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

The corresponding output is given by

31.837 105 0.0023 23.04 4eff

o s

GFV V V mV

−∈ ×≈ = × = =



2) Diaphragm/Bellows type transducer:

Pressure signal is converted to a displacement in the case of

diaphragm/bellows type pressure gage. The diaphragm or bellows acts as a

spring element that undergoes a displacement under the action of the

pressure. Schematic of diaphragm and bellows elements are shown

respectively in Figure 79 (a) and (b).

(a)

(b)

Figure 79 Schematic of diaphragm and bellows type pressure gages that use

LVDT for displacement measurement

In the above figure a Linear Variable Differential Transformer (LVDT) is used

as a displacement transducer. The working principle of LVDT will be

discussed later on. In the case of the diaphragm type gage one may also use

a strain gage for measuring the strain in the diaphragm by fixing it at a

suitable position on the diaphragm (Figure 80).

P1 P2

Diaphragm

LVDT

LVDT

Bellows

P2

P1



Figure 80 Diaphragm gage with strain gage for Displacement measurement

In the case of the bellows transducer the relationship between the pressure

difference and the displacement is given by

( )1 2p p AK

δ−

= (86)

Here δ is the displacement of the bellows element, A is the inside area of the

bellows and K is the spring constant of the bellows element. In the case of

the diaphragm element the following formula from strength of materials is

useful:

( ) ( )2 2 2

1 2 23

3 ( )( ) 1

16p p a r

rE t

δ ν− −

= − (87)

In the above formula δ is the deflection at radius r, ‘a’ is the radius of the

diaphragm, t is its thickness, E and ν are respectively the Young’s modulus

and Poisson ratio of the material of the diaphragm. This formula assumes

that the displacement of the diaphragm is small and the deformation is much

smaller the elastic limit of the material. The perimeter of the diaphragm is

assumed to be rigidly fixed. The force is assumed to be uniformly distributed

over the surface of the diaphragm. It is seen that the maximum deflection

takes place at the center of the diaphragm

p1

Strain gage

p2 a

r



Figure 81 Strain gages for mounting on a diaphragm gage

(Figures from Vishay Intertechnology, Inc., Manufacturers of Diaphragm pressure gages) Terminals are provided for external connections and for compensating resistors. Different

strain gage elements of the bridge respond to radial (∈R) and tangential (∈T) strains Since the strain gage occupies a large area over the diaphragm it is

necessary to integrate the strain of the element over its extent. The

displacement given by Equation 87 is normal to the plane of the diaphragm.

Because of this displacement the diaphragm undergoes both radial as well as

circumferential strains. As indicated in Figure 81 strain gage elements

respond to both the radial as well as tangential (circumferential) strains. The

numbers represent the leads and the strain gages are connected in the form

of a full bridge. The ratio of output to input is then given by the approximate

formula (from Vishay Intertechnology, Inc., Manufacturers of Diaphragm

pressure gages)

( )( )2 21 2 3

10.75 10 /o

s

p pV a mV VV E t

ν− − ⎛ ⎞= ×⎜ ⎟⎝ ⎠

(88)

The above formula assumes that the diaphragm is perfectly rigidly held over

its perimeter. This is achieved by making the diaphragm an integral part of

the body of the pressure transducer. It is also assumed that the sensitivity of

the strain gage is 2. While measuring transient pressures it is necessary that



the transients do not have significant components above about 20% of the

natural frequency (fn) of the diaphragm given by the following expression.

( )2 2

0.4691n

t Efa ρ ν

=−

(89)

In the above expression ρ is the density of the material of the diaphragm in

kg/m3. The diaphragm is again assumed to be rigidly clamped along its

periphery.



Example 29

Steel diaphragm of 15 mm diameter is used in a pressure transducer to

measure a maximum possible pressure of 10 MPa. The transducer is

required to give an output of 3 /mV V at the maximum pressure. Use

the following data to determine the thickness of the diaphragm. Also

determine the maximum out of plane deflection of the diaphragm and

the its natural frequency.

Young’s modulus 207E GPa= Poisson ratio 0.285ν = The gage is constructed as shown in Figure 81.

From the given data we have:

3 /o

s

V mV VV

= ; 15 0.0752

a mm m= = ; ( ) 71 2 10 10p p MPa Pa− = =

We use Equation 88 to obtain the diaphragm thickness as

( )( )

( )

2 31 2

7 2 3

11

0.75 1 10 1

0.75 10 1 0.285 10 10.0752.07 10 3

0.000593 0.593

o

s

p pt a

E VV

m mm

ν− − ×=

× × − ×= × ×

×= =

The maximum out of plane deflection is obtained by putting 0r = in

Equation 87.

( ) ( )7 4 2

11 3

3 10 0.0075 1 0.2850 0.0001263 0.126

16 2.07 10 0.000593r m mmδ

× × × −= = = =

× × ×

The ratio of the maximum deflection to the diaphragm thickness is

( )0 0.000126 0.2130.000593t

δ= =

This ratio is less than 0.25 and hence is satisfactory.

The natural frequency is given by



( )9

2 2

0.469 0.000593 207 10 265220.0075 7830 1 0.285nf Hz× ×

= =× −

The sensor is useful below about 5 kHz Diaphragm/Bellows type gage with LVDT: As indicated earlier a diaphragm/bellows element may be used to measure

pressure by directly measuring the small displacement very accurately. This

may be done by using a Linear Variable Differential Transformer (LVDT) as

shown in Figure 63. A schematic of LVDT is shown in Figure 82.

Figure 82 Schematic of an LVDT

Figure 83 Operation of the LVDT

Secondary 1

Vo=V1-V2

Secondary 2

V1

V2 Primary

Co

re

Movement

Secondary Coil 1 Secondary Coil 2

Primary Coil Ferrite Core

SS = Support Spring

S S



LVDT consists of three coaxial coils with the primary coil injected with an ac

current. The coils are mounted on a ferrite core carried by a stainless steel

rod. The two secondary coils are connected in opposition as shown in Figure

83.

The stainless steel rod is restrained by springs as indicated. One end of the

rod is connected to the diaphragm/bellows element as indicated in Figure 79.

When the ferrite core is symmetrical with respect to the two secondary coils

the ac voltages induced by magnetic coupling across the two secondary

coils are equal. When the core moves to the right or the left as indicated the

magnetic coupling changes and the two coils generate different voltages and

the voltage difference is proportional to the displacement (see Figure 84).

Figure 84 Output of an LVDT excited by alternating current supply. The phase information is with respect to the phase of the supply.

Figure 85 Schematic of a typical LVDT signal conditioner

Demodulator

Demodulator

Sum

Difference

Divider Gain A

Primary Input

Output DC



We describe the typical LVDT circuit shown in Figure 85. The primary input

voltage is 1 – 3 V rms. Full scale secondary voltage is between 0.1 – 3 V

rms. Gain is set between 2 – 10 to increase the output voltage to ± 10 V. The

two secondary coils are connected in opposition and sum and difference

signals are obtained from the two coils, after demodulating the two voltages.

The ratio of these is taken and is output with a gain A as indicated in the

figure. The output is independent of the gain settings of the sum and

difference amplifiers as long as they have the same gain.

Table 16 Common specifications for commercially available LVDT

Input: Power input is a 3 to 15 V (rms) sine wave with a frequency

between 60 to 20,000 Hz (the two most common signals are 3 V, 2.5 kHz and 6.3 V, 60 Hz).

Stroke: Full-range stroke ranges from ±125 µm to ±75 mm Sensitivity: Sensitivity usually ranges from 0.6 to 30 mV per 25 µm

under normal excitation of 3 to 6 V. Generally, the higher the frequency the higher the sensitivity.

Nonlinearity: Inherent nonlinearity of standard units is on the order of 0.5% of full scale.

In Example 29 the diaphragm undergoes a maximum deflection of 0.126 mm

or 136 μm. As Table 16 shows we can certainly pick a suitable LVDT for this

purpose.



3) Capacitance diaphragm gage:

Figure 86 Capacitance pressure transducer

Figure 86 shows the schematic of a capacitance type pressure transducer.

The gap between the stretched diaphragm and the anvil (remains fixed

because of its mass) varies with applied pressure due to the small

displacement of the diaphragm. This changes the capacitance of the gap.

The theoretical basis for this is derived below.

Consider a parallel plate capacitor of plate area A cm2 and gap between

plates of x cm. Let the gap contain a medium of dielectric constant κ. The

capacitance of the parallel plate capacitor is then given in pico-Farads (1 pF =

10-12 Farad) by

0.0885 AC pFx

κ= (90)

The dielectric constants of some common materials are given in Table 17.

Table 17 Dielectric constants of some common materials

Material κ Material κ Vacuum 1 Plexiglass® 3.4 Air (1 atmosphere) 1.00059 Polyethylene 2.25 Air (10 atmospheres)

1.0548 PVC 3.18

Glasses 5 – 10 Teflon 2.1 Mica 3 – 6 Germanium 16 Mylar 3.1 Water 80.4 Neoprene 6.7 Glycerin 42.5

p

Diaphragm

Variable Gap

Anvil



Logarithmic differentiation of Equation 90 shows that

20.0885C x ACx x

κΔΔ = − = − (91)

As an example consider a capacitance type gage with air as the medium,

21A cm= , 0.3 0.03x mm cm= = . We then have 0.0885 1 2.950.03

C pF×= = . The

sensitivity defined by the partial derivative of C with respect to x

is 2.95 98.3 /0.03

C CS pF cmx x

∂= = − = − = −∂

.

Bridge circuit for capacitance pressure gage: Schematic of a bridge circuit for a capacitance pressure gage is shown in

Figure 87. The circuit is driven by alternating current input at high frequency

and coupled using a transformer, as shown in the figure.

Figure 87 Bride circuit for capacitance pressure gage

P

Pref

AC Output

Diaphragm



Electrical input and output Electrical input and output of pressure transducersof pressure transducers

• Standard input:–5-24 V DC for amplified voltage

output transducers–8-30 V DC for 4-20 mA current

output• Standard output:

–0-5 or 0-10 V DC output (used in industrial environments)

–4-20 mA current output (pressure transmitters)

Figure 88 Summary of typical electrical input and output values

Either a balanced or unbalanced mode of operation of the bridge circuit is

possible. In the balanced mode of operation the variable capacitor is adjusted

to bring the bridge back to balance. The position of a dial attached to the

variable capacitor may be marked in pressure units. Alternately the output

may be related to the pressure by direct calibration. Figure 88 indicates the

typical electrical input output pairs for pressure transducers. In the current

output type of instrument 4 mA corresponds to 0 pressure and 20 mA

corresponds to full scale (dependent on the range of the pressure transducer).

Current transmitters of this type will be discussed later. The full scale may be

a few mm of water column up to pressures in the MPa range. The scale is

linear over the entire range of the transducer. Note that this type of gage may

also be used for measurement of pressures lower than the atmospheric

pressure.

Measurement of transient pressures: In continuation of what we have said about transient pressures, we note

the following points:

• Transient behavior of pressure measuring instruments is needed

to understand how they will respond to transient pressures.



• Measurement of transient or time varying pressure requires a

proper choice of the instrument and the coupling element

between the pressure signal and the transducer.

• Usually process pressure signal is communicated between the

measurement point and the transducer via a tube.

• The tube element imposes a “resistance” due to viscous friction.

• Viscous effects are normally modeled using laminar fully

developed flow assumption (discussed earlier while considering

the transient response of a U tube manometer).

Before we proceed with the analysis we shall look at a few general things.

These are in the nature of comparisons with what has been discussed earlier

while dealing with transients in thermal systems and in the case of the U tube

manometer. We shall keep electrical analogy in mind in doing this.

Thermal system: Thermal Capacity C (J/K) = Product of mass (M, kg) and specific heat (c, J/kg

K). With temperature T representing the driving potential, he representing the

enthalpy the specific heat capacity is given by ehcT

∂=∂

. Then, we

have, ehC Mc VT

ρ ∂= =

∂. Thermal resistance is given by 1R

hS= where h is the

heat transfer coefficient and S is the surface area. The time constant is given

by McRChS

τ = = (using analogy with an electrical circuit).

Pressure measurement in a liquid system: Driving potential is the head of liquid h (m). Volume of liquid V (m3) replaces

enthalpy in a thermal system. Thus the capacitance is dVCdh

= and has units

of m2. We have already seen that the resistance to liquid flow is defined



through the relation dpRdm

= − with p standing for pressure and m standing for

the mass flow rate. Again the time constant is dV dpRCdh dm

τ = = − .

Gas system: Potential for bringing about change is the pressure p. The mass of the gas m

replaces the enthalpy in the thermal system. The capacitance is given

by dmCdp

= . With m being given by the product of density and volume, we

have ( )dm d VCdp dp

ρ= = . The capacitance depends on the process that the gas

undergoes while the pressure transducer responds to the transient. If we

assume the process to be polytropic, then Constantn

pρ

= , where n has an

appropriate value ≥ 1. If, for example, the volume of the gas is held constant,

we have dC Vdpρ

= . For the polytrope, logarithmic differentiation

gives 0dp dnp

ρρ

− = or ddp npρ ρ= . With this the capacitance becomesC V

npρ

= .

Assuming the gas to be an ideal gas gp R Tρ= and hence the capacitance

isg

VCnR T

= . The resistance to flow is again given by the Hagen-Poiseuelle

law as in the case of a liquid. The time constant is then given byg

VRnR T

τ = .

Transient response of a bellows type pressure transducer: The bellows element may be considered as equivalent to a piston of area A

(in m2) spring arrangement as shown in Figure 89. Let the bellows spring

constant be K (in N/m). Let the displacement of the bellows element be b (in

m). Let the bellows gage be connected to a reservoir of liquid of density ρ

(kg/m3) at pressure P (in Pa). At any instant of time let p (in Pa) be the



pressure inside the bellows element. The displacement of the spring is then

given by the relation

ForceSpring constant

pAbK

= = (92)

Figure 89 Transient in a bellows type gage Let the connecting tube between the reservoir and the gage be of length L (in

m) and radius r (in .m). If the pressure P is greater than the pressure p, the

liquid will flow through the intervening tube at a mass flow rate given

by dbm Adt

ρ= . We know from the definition of connecting tube resistance that

this should equal P pmR−

= . All we have to do is to equate these two

expressions to get the equation governing the transient.

KP bdb P p AAdt R R

ρ−−

= = (93)

The latter part of the equality follows from Equation 92. The above equation

may be rearranged in the form

2A R db Ab PK dt Kρ⎛ ⎞

+ =⎜ ⎟⎝ ⎠

(94)

`We see that the transient of a bellows type gage is governed by a first order

ordinary differential equation with a time constant2A RKρτ = . Hence the

bellows type gage is a first order system just like the first order behavior

L

Pressure P

Coupling Element

p

Displacement Element

Bellows (Spring)

Fixity

2r



exhibited by a first order thermal system. We infer from the above that the

capacitance of a bellows type gage is given by2ACKρ

= .



Example 30

A bellows pressure gage of area 1 cm2, spring constant K = 4.4 N/cm is

connected by a 2.5 mm ID tubing that is 15 m long in a process

application. Measurement is of pressure of water at 30°C. Determine the

time constant of this gage.

The given data for the gage:

2 4 21 10 , 4.4 / 440 /A cm m K N cm N m−= = = = The given data for the connecting tube:

2.5 1.25 0.00125 , 152 2IDr mm mm m L m= = = = =

Properties of water required are taken from tables. These are

3 6 2995.7 / , 0.801 10 /kg m m sρ ν −= = ×

The tube resistance is calculated as

( )6

174 4

8 8 0.801 10 15 1.2532 100.00125

LR m srνπ π

−−× × ×

= = = ××

The capacitance is given by

( )2428 2

10 995.72.26 10

440AC s mKρ

−−

×= = = ×

The time constant is then given by

7 81.2532 10 2.26 10 0.284RC sτ −= = × × × =



Force balancing element for measuring pressure:

Figure 90 Transient in a force balance transducer Figure 90 shows the arrangement in a force balance element. It consists of

a diaphragm gage that is maintained in its zero displacement condition by

applying a force that balances the pressure being measured. The required

force itself is a measure of the pressure. Let us assume that the pressure

being measured is that of a gas like air. If the diaphragm displacement is

always in the null the volume of the gas within the gage is constant. If the

pressure P is greater than pressure p mass flow takes place from the

pressure source in to the gage to increase the density of the gas inside the

gage. The time constant for this system is just the one discussed earlier and

is given by

g

VRnR T

τ = (95)

In case the instrument is not the null type (where the diaphragm displacement

is zero at all times), the capacity should include both that due to density

change and that due to displacement. Since the displacement is very small, it

is appropriate to sum the two together and write

2

g

A VCK nR Tρ

= + (96)

The above is valid as long as the transients are small so that the changes in

density of the fluid are also small. In most applications this condition is

satisfied.

Force p

Pressure P

L

2r



Example 31

A bellows type pressure gage of effective area A = 0.25 cm2, spring

constant K = 100 N/m is connected by a 1.5 mm ID 10 m long stainless

tubing in a process application. The volume of the effective air space

in the bellows element is V = 2 cm3. Determine the time constant if the

pressure measurement is a pressure around 5 bars and the

temperature is 27°C.

We assume that the pressure hovers around 5 bars during the

transients. The air properties are taken from appropriate tables.

Air properties:

55 5 10 , 27 300 , 287 /gp bars Pa T C K R J kg K= = × = ° = =

The density of air is calculated as

535 10 5.807 /

287 300g

p kg mR T

ρ ×= = =

×

The dynamic viscosity of air is (from table of properties)

51.846 10 /kg m sμ −= × The kinematic viscosity is hence given by

56 21.846 10 3.179 10 /

5.807m sμν

ρ

−−×

= = = ×

The capacity is calculated based on Equation 96

( )242 611 2

0.25 10 5.807 2 10 5.289 10100 1.4 287 300g

A VC m sK nR Tρ

− −−

× × ×= + = + = ×

× ×

In the above we have assumed a polytropic index of 1.4 that

corresponds to an adiabatic process.

The fluid resistance is calculated now as

68 1

44

8 8 3.179 10 10 2.559 10 ( )0.0015

2

LR m srνπ

π

−−× × ×

= = = ×⎛ ⎞×⎜ ⎟⎝ ⎠

The time constant of the transducer is then given by 8 112.559 10 5.289 10 0.0135RC sτ −= = × × × =



Example 32

Rework Example 28 assuming that the gage operates in the null mode.

The capacity is given by

611 22 10 1.659 10

1.4 287 300g

VC m snR T

−−×

= = = ×× ×

In the above we have assumed a polytropic index of 1.4 that

corresponds to an adiabatic process.

The fluid resistance is calculated now as

68 1

44

8 8 3.179 10 10 2.559 10 ( )0.0015

2

LR m srνπ

π

−−× × ×

= = = ×⎛ ⎞×⎜ ⎟⎝ ⎠

The time constant of the transducer is then given by

8 112.559 10 1.659 10 0.0042RC sτ −= = × × × =

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